The expected value of the distance is the sum of the probabilities that the distance is greater than $d$ for $d=0,1,2, ...$. Unfortunately, it's hard to determine the exact value of this probability. Whether this is greater than $0$ determines whether a code of that size exists.
There are some easy bounds on the probabilities which translate to bounds on the expected minimum Hamming distance. Take any ordering of the points, and check whether each point is consistent with the previous points. As an upper bound, the balls of radius $\lfloor d/2 \rfloor$ about the previous points must be disjoint, and the current point must not be contained in these balls. As a lower bound, each previous choice rules out at most a ball of radius $d$ new points.
Let $B(L,d,q) = \sum_{i=0}^d {L \choose i}(q-1)^i$ be the size of the ball of radius $d$ with an alphabet of size $q$. Let $Pr(L,d,q,P)$ be the probability that $P$ random points are at distances strictly greater than $d$ from each other.
$$ \prod_{i=1}^{P-1} \max(0,1-\frac{iB(L,d,q)}{q^L}) \le Pr(L,d,q,P) \le \prod_{i=1}^{P-1} \max(0,1-\frac{iB(L,\lfloor d/2 \rfloor, q)}{q^L}).$$
The upper bound on the expected minimum Hamming distance from summing the probabilities is precisely twice the lower bound, so these bounds get you within a factor of $2$. Actually, that upper bound on the probabilities is so weak that a better bound in many situations is $Pr(L,d,q,P) \le (1-B(L,d,q)/q^L)^{P-1}$. You get that by saying that the second and subsequent points have to avoid the ball of radius $d$ about the first point.
For example, suppose you take $q=2, L=20, P=10.$ The lower bound on the expected minimum Hamming distance is $5.10246$. If you use the second upper bound when it is better, the upper bound on the expected minimum distance is $6.71209.$
The only rigorous bound I am aware of is due to Gonzalo Navarro*
$$c\geq 1-{\rm e}/\sqrt{\sigma},$$
for an alphabet of $\sigma$ characters. Obviously, for the binary string ($\sigma=2$) this bound is ineffective. Navarro also mentions a large-$\sigma$ conjecture $c=1-1/\sqrt{\sigma}$, which for the binary string would give $c=0.2929$, quite close to your numerical finding.
*G. Navarro, A guided tour to approximate string matching (2001)
Best Answer
This is not a complete answer and may not tell you anything you don't already know, but it's too long for a comment.
You can get an estimate on the number of words at an edit distance of $k$ from a given word of length $n$ by using the arguments in the proof of Lemma 2.6 in this paper. (The lemma is on page 10 and its proof is on page 28.)
The idea is that if a word $w$ is at an edit distance of $k$ from a word $v$, then one can get from $v$ to $w$ via the following steps:
Step 1 gives a word of length $n+k$ with $k$ occurrences of the symbol 'e', so there are ${n+k\choose k}$ possibilities after this step. Then step 2 gives at most $3^k$ possible words for each of those possibilities, so for a fixed word $v$ of length $n$ one obtains the bound $$ \#\{w \mid \text{edit distance from $v$ to $w$ is $k$}\} \leq 3^k {n+k \choose k}. $$ For small $k$ this is a reasonable bound; in particular when $k\ll n$ this is not much larger than ${n\choose k}$, the number of words a Hamming distance of $k$ away. The problem is that for larger values of $k$ the procedure described above succumbs to a lot of overcounting, so it's not clear what the actual bound should be.